Betty goes to the store to get flour and sugar.  The amount of flour she buys, in pounds, is at least 6 pounds more than half the amount of sugar, and is no more than twice the amount of sugar.  Find the least number of pounds of sugar that Betty could buy.
Solution: Suppose Betty buys $f$ pounds of flour and $s$ pounds of sugar.  From the problem, we have $f \ge 6+s/2$ and $f \le 2s$.  Putting these together, we have $2s \ge f \ge 6 +s/2$. The expression on the left end of this inequality chain must therefore be greater than or equal to the $6+s/2$ on the right, so \[2s \ge 6 + s/2 \implies 3s/2 \ge 6 \implies s\ge \boxed{4}.\]